Parabola Analysis: Notes and Practice Questions Key Concepts 1. General Form of a Parabola: f (x) = ax2 + bx + c - a: Determines the direction of the parabola’s opening. - If a > 0, the parabola opens upwards. - If a < 0, the parabola opens downwards. - b: Affects the axis of symmetry and vertex. - c: The y-intercept (point where the parabola crosses the y-axis). 2. Axis of Symmetry: The axis of symmetry is the vertical line that passes through the vertex of the parabola. It can be calculated using: x=− b 2a 3. Vertex: The vertex is the highest or lowest point of the parabola (depending on whether it opens up or down). Its coordinates are: b b Vertex: − , f − 2a 2a 4. Y-Intercept: The y-intercept is found by setting x = 0 in the equation of the parabola: y=c 5. X-Intercepts: The x-intercepts are the points where the parabola crosses the x-axis. These are found by solving: ax2 + bx + c = 0 Using the quadratic formula: x= −b ± √ b2 − 4ac 2a Practice Questions and Solutions Example 1: Compare Two Parabolas Given f (x) = x2 − 5x − 70 and g(x) = x2 +7x−71 , 2 1. Do the parabolas open up or down? • f (x) : a = 1 > 0, so it opens up. • g(x) : a = 1 2 > 0, so it opens up. 2. What are their axes of symmetry? −5 • f (x) : x = − 2(1) = 25 . 7 • g(x) : x = − 2·21 = − 27 . 2 3. What are their y-intercepts? 1 answer the following: • f (x) : y = −70. • g(x) : y = −35.5. 4. What are their x-intercepts? • f (x) : x = 10, x = −7. • g(x) : x = 10, x = −7. Example 2: Ball Bounce Problem (Geometric Sequence) A ball is dropped from a height of 500 meters. After each bounce, the height decreases by a factor of 0.8. Write the equation for the height after n bounces and calculate the height on the 6th bounce. Solution: • General formula: hn = 400 · (0.8)n−1 . • Height after the 6th bounce: h6 = 400 · (0.8)5 = 400 · 0.32768 = 131.07 meters. Example 3: Savings Problem (Arithmetic Sequence) If you have deposited $115 in a savings account and added $35 weekly. Write a formula for the total deposited after n weeks and calculate the total after 30 weeks. Solution: • Formula for the amount deposited in the n-th week: an = 115 + (n − 1) · 35 = 35n + 80. • Total deposited after 30 weeks: 30 · (115 + 1130) 2 = 15 · 1245 = 18, 675. S30 = 2